Feel nervous about your number-crunching math skills? Try to improve by practicing our Consulting Case Interview Fast Math Drills quiz. Note: No calculator is allowed during the quiz, and you may use a pencil and a piece of scratch paper. Try your best to answer all 10 questions in 10 minutes. The correct answers will be shown at the end of the quiz.
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how does question 10 work? I may have misunderstood the question?
Hi, Can anybody explain how to work out question 3? Thanks
Suppose the lawn has an area of 60 m^2,
it takes John exactly 30 minutes to rake the lawn, John’s speed is 60 m^2 / 30 min = 2 m^2/min,
it takes Todd exactly 60 minutes to rake the lawn, Todd’s speed is 60 m^2 / 60 min = 1 m^2/min,
if they work together, their total speed is 3 m^2/min, it will take them 60 m^2 / (3 m^2/min) = 20 min
Thanks
With the work formula is way easier:
1/30 + 1/60 = 1/T,
then T = 20 min
5. A packer of boxes in a Target warehouse is packing small radios into larger boxes that measure 25 in. by 43 in. by 62 in. If the measurement of each radio is 7 in. by 6 in. by 5 in., then at most how many radios can be placed in the box?
A. 280
B. 288
C. 300
D. 315
E. 317
(25/5) * (62/6) * (43/7) = 5*10*6 = 317 and not 300…
You have to round down (62/6) to 10, and (43/7) to 6.
How do you decide what to divide by what? The correct way to do this, as Anita said, would be to divide total volume inside the box , by the total volume for each radio. Hence (25*43*62) divided by (7*6*5) which gives 317 radios. If we follow the solution as given in the key, then it doesn’t make sense. It is unclear why 25 is divided by 5 and not by 7 or 6… and so on. For example : (25/5)*(43/6)*(62/7) would give us 5*7*8 which is 280 (also one of the options). So please provide the logic provided to arrive at the answer given. Thank you.
You have to find the greatest value by dividing dimensions of box to that of the radios. A radio can not be squeezed or twisted to fit into the box, so only whole numbers of radios will fit into the boxes. Therefore the values are 300 and not 317. This problem includes both logical reasoning as well as maths.
Can anyone clarify the answer? obviously it is not 300, since you could easily pack more boxes than that, but due to the fact you cannot squeeze the box, you could not simply divide the total volume by unit volume. I got 310 as the answer, here is how I got it: (25/5) * (43/6) * (62/7)=5*7*8=280, and you have a layer of dimension 25*43 with height 62-7*8=6, then again you divide this volume (25/5)*(6/6)*(43/7)=30, the total is 280+30=310. Is this more efficient way to use the space? maybe, if you don’t align each box in an organized way.
To Cookie: your solution is brilliant! In real business, if a retailer can reduce their shipping costs by (1 – 300/310 = 3%), it is certainly worth trying. Thank you for your input, we have changed the correct answer to 310
@Cookie
Nope you got the most efficient solution, well done.
Total box volume: 6665 (25*43*62)
Total space taken by 310 boxes: 310*7*6*5=6510
Volume left over: 155
Volume of a single box: 6*7*5= 210
210>155
I’m actually a bit confused with cookies’ explanation….
hope this help to clear it a bit……
well… cookies’ right, but I don’t agree with consulting 101
if we make a matrix:
5in 6in 7in
———————————————————————————————————–
25 | stack 5 with 0in left | stack 4 with 1in left | stack 3 with 4in left
43 | stack 8 with 3in left | stack 7 with 1in left | stack 6 with 1in left
62 | stack 12 with 2in left | stack 10 with 2in left | stack 8 with 6in left
cookie’s solution would be to align the 5in side to the 25in side = 5 boxes
align the 6in side to the 43in side = 7 boxes
align the 7in side to the 62in side = 8 boxes
this would fit 280 boxes
there will be 6in leftover on the 62in side which would leave us a 25×43 space with boxes with 6×7 space
which would best fit 5 boxes (5in side to the 25in with 0in left) by 6 boxes (7in side to the 43in space with 1in left)
so there’s the additional 30 boxes….. which makes the total 310 boxes……
but IMO, the flawed one is the question, as it doesn’t fulfill the purpose of “fast math drills” which should focus more on the quick and precise-ish estimation/educated guesses….. but to came up with the 310 boxes you need to drill and sit on the problem looking for alternative solution (in this case putting boxes in different position) which IMO is more appropriate for an essay/discussion question rather than multiple-choice like this one…..
so IMO 310 IS the maximum amount of boxes you can fit, but 300 (5 by 10 by 8 boxes on 25in, 62in and 43in sides) is the more appropriate answer for this type of question and this type of quiz
@consultingcase101, the optimal answer is actually 315, not 300 nor 310.
Because 5 divides perfectly into 25, ignore that dimension for now. Just pack 6×7 rectangles into 43×62, then we’ll stack it five layers high to fill the 25″.
How to fill 43×62 with 7×6 blocks? Think of this as 43 across, 62 down.
Place a *single* row of (6) 7×6 blocks across. Now you have 43×56 remaining (plus a 1×6 space at the right edge).
Add a *single* column of (9) 7×6 blocks down. Now you have 36×56 remaining (plus a 7×2 space at left bottom edge).
Fill the remaining space perfectly with blocks turned to the 6×7 orientation: six across x eight down: (48).
This gives you 6+9+48 or (63) blocks that can fit into 43×62.
Remember the third dimension: the blocks are 5″ high. The enclosure is 25″ high.
So stack up five levels and you get 63×5, or (315) blocks!
By the way, total empty space for each layer is (6x1x5 + 7x2x5) or 100 cub.in. Across 5 layers, this leaves 500 cub.in., which theoretically fit 2 additional blocks. But you knew that already since 25x43x62=66650, while 315 x 5x6x7 = 66150.
There is a mistake on question number 2:
470/35,000 = 0.00134 = 1.34%
The right answer would be 470/35,000 = 0.0134 = 1.34%
Thanks for the catch, corrected.